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Saturday, 30 November 2013

The trace distance between pure states

In this post, the relationship between two distance measures on density operators will be analyzed for a few specific cases. Namely, the trace distance and the Euclidean distance of two pure states will be studied.

The trace norm is defined as $||M||_{Tr}=Tr(\sqrt{M^\dagger M})$, and
the trace distance between two density operators $\rho$ and $\sigma$ is given by
$||\rho-\sigma||_{Tr}$.

Let's calculate an expression for the trace distance between $\ket{0}$ and $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ as a function of $\theta$.

As another example, lets now calculate an expression for the Euclidean distance between the two points in the Bloch sphere that correspond to the pure states $\ket{0}$ and $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$.

In the Bloch sphere representation, a general density matrix can be expressed as\[\frac{1}{2}(I+c_xX+c_yY+c_zZ),\]where $X$, $Y$, $Z$ are the Pauli matrices, and the vector $(c_x,c_y,c_z)$ gives the coordinates of the density matrix on the Bloch sphere. In such a representation, $\rho$ and $\sigma$ can be expressed as follows:\[\begin{align*}\rho=&\frac{1}{2}(I+Z),\\\sigma=&\frac{1}{2}(I+2\cos(\theta)\sin(\theta)X+\cos(2\theta)Z),\end{align*}\]so that the coordinate vectors for the Block sphere representations of $\rho$ and $\sigma$ are given by, respectively,\[\begin{align*}v_{\rho}=&(0,0,1) \\v_{\sigma}=&(2\cos(\theta)\sin(\theta),0,\cos(2\theta)).\end{align*}\] Then the Euclidean distance between $\rho$ and $\sigma$ on the Block sphere is given by \[\begin{align*} ||v_\rho-v_\sigma||_2&=\sqrt{(-2\cos(\theta)\sin(\theta))^2+(1-\cos(2\theta))^2}\\ &=\sqrt{4\cos^2(\theta)\sin^2(\theta)+4\sin^4(\theta)} \\ &=\sqrt{4\sin^2(\theta)(\cos^2(\theta)+\sin^2(\theta))}\\ &=\sqrt{4\sin^2(\theta)}\\ &=2|\sin(\theta)|. \end{align*}\]

Hence, $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=2|\sin(\theta)|$, and the two distance measures agree.

Now we'll repeat the calculations done above for these two states: $\rho=\begin{pmatrix} 1/2 &0 \\ 0 & 1/2 \end{pmatrix}$ and the pure state $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ as a function of $\theta$.

As calculatedabove, the density matrix corresponding to the state $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ is given by\[ \sigma:=\begin{pmatrix}\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\sin^2(\theta) \end{pmatrix}\]

Calculating the relevant matrices in order to determine the trace distance $||(\rho-\sigma)||_{Tr}=Tr(\sqrt{(\rho-\sigma)^2})$ yields

Now, to calculate the Euclidean distance between the two states in the Block sphere representation observe that

\[ \begin{align*}\rho=&\frac{1}{2}I,\\\sigma=&\frac{1}{2}(I+2\cos(\theta)\sin(\theta)X+\cos(2\theta)Z),\end{align*}\]so that the coordinate vectors for the Block sphere representations of $\rho$ and $\sigma$ are given by, respectively,\[\begin{align*}v_{\rho}=&(0,0,0) \\v_{\sigma}=&(2\cos(\theta)\sin(\theta),0,\cos(2\theta)).\end{align*}\]

The Euclidean distance is then given by \[ \begin{align*} ||v_\rho-v_\sigma||_2&=\sqrt{(-2\cos(\theta)\sin(\theta))^2+(-\cos(2\theta))^2}\\ &=\sqrt{4\cos^2(\theta)\sin^2(\theta)+\cos^2(2\theta)} \\ &=\sqrt{4\cos^2(\theta)\sin^2(\theta)+(\cos^2(\theta)-\sin^2(\theta))^2}\\ &=\sqrt{2\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\sin^4(\theta))^2}\\&=\sqrt{2\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\sin^4(\theta))^2}\\ &=\sqrt{(\cos^2(\theta)+\sin^2(\theta))^2}\\ &=\sqrt{1^2}\\ &=1, \end{align*}\] and so the trace distance of $\rho$ and $\sigma$ agrees with their Euclidean distance on the Block sphere since $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=1$.